hdu 1394 树状数组求逆序数

          以前用过线段树求逆序数，这次想用树状数组试一下，悲催的是想了好久才想明白。。。。看来对树状数组还是不够了解啊。纠结。。。。题目：

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2413    Accepted Submission(s): 1492

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

Output

For each case, output the minimum inversion number on a single line.

 

Sample Input

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

ac代码：

#include <iostream>#include <cstdio>#include <string.h>int num[5005];int total[5005];int n;int lowbit(int x){  return x&(-x);}void update(int x){while(x>0){  total[x]++;  x-=lowbit(x);}//printf("***\n");}int add(int x){int ss=0;x+=1;while(x<=n){  ss+=total[x];  x+=lowbit(x);}return ss;}int main(){  //int n;  while(~scanf("%d",&n)){    memset(num,0,sizeof(num));memset(total,0,sizeof(total));int sum=0;for(int i=0;i<n;++i){  scanf("%d",&num[i]);  update(num[i]+1);  sum+=add(num[i]+1);}//printf("sum===%d\n",sum);int k=0,s=sum;for(int i=0;i<n;++i){  k=sum-num[i]+(n-1-num[i]);  sum=k;  //printf("k==%d   s==%d\n",k,s);  if(k<s)  s=k;}printf("%d\n",s);  }  return 0;}